সৃজনশীলঃ ‘a=sec⁡θ−tan⁡θ যেখানে 0<θ<π2.…’ চট্টগ্রাম বোর্ড ২০২৩ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $a=\sec \theta -\tan \theta$

$\therefore$ ডানপক্ষ $=\frac{1-a}{1+a}=\frac{1-\sec \theta +\tan \theta }{1+\sec \theta -\tan \theta }$

$=\frac{1-\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }}{1+\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }}=\frac{\frac{\cos \theta -1+\sin \theta }{\cos \theta }}{\frac{\cos \theta +1-\sin \theta }{\cos \theta }}$

$=\frac{\cos \theta -1+\sin \theta }{\cos \theta +1-\sin \theta }=\frac{\sin \theta (\cot \theta -\mathrm{cosec}\theta +1)}{\sin \theta (\cot \theta +\mathrm{cosec}\theta -1)}$

$=\frac{\cot \theta -\mathrm{cosec}\theta +1}{\cot \theta +\mathrm{cosec}\theta -1}$

$=\frac{(\cot \theta -\mathrm{cosec}\theta )+(\mathrm{cosec}{}^2\theta -{\cot }^2\theta )}{\cot \theta +\mathrm{cosec}\theta -1}$

$=\frac{(\cot \theta -\mathrm{cosec}\theta )+(\mathrm{cosec}\theta +\cot \theta )(\mathrm{cosec}\theta -\cot \theta )}{\cot \theta +\mathrm{cosec}\theta -1}$

$=\frac{(\mathrm{cosec}\theta -\cot \theta )(-1+\mathrm{cosec}\theta +\cot \theta )}{\cot \theta +\mathrm{cosec}\theta -1}$

$=\frac{(\mathrm{cosec}\theta -\cot \theta )(\cot \theta +\mathrm{cosec}\theta -1)}{(\cot \theta +\mathrm{cosec}\theta -1)}$$=\mathrm{cosec}\theta -\cot \theta$

= বামপক্ষ

$\therefore \mathrm{cosec}\theta -\cot \theta =\frac{1-a}{1+a}$ (প্রমাণিত)

দেওয়া আছে, $\sec \theta -\tan \theta =a$

এবং $\mathrm{a}=\frac{1}{\sqrt{3}}$

বা, $\sec \theta -\tan \theta =\frac{1}{\sqrt{3}}$

বা, $\sqrt{3}\sec \theta -\sqrt{3}\tan \theta =1$

বা, $\sqrt{3}\sec \theta =1+\sqrt{3}\tan \theta$

বা, $(\sqrt{3}\sec \theta {)}^2=(1+\sqrt{3}\tan \theta {)}^2$ [বর্গ করে]

বা, $3{\sec }^2\theta =1+2\sqrt{3}\tan \theta +3{\tan }^2\theta$

বা, $3+3{\tan }^2\theta =1+2\sqrt{3}\tan \theta +3{\tan }^2\theta$

বা, $2\sqrt{3}\tan \theta =2$

বা, $\tan \theta =\frac{2}{2\sqrt{3}}$

বা, $\tan \theta =\frac{1}{\sqrt{3}}$

বা, $\tan \theta =\tan \frac{\pi }{6}\therefore \theta =\frac{\pi }{6}\cdot$

$[\because 0<\theta <\frac{\pi }{2}]$

∴ নির্ণেয় মান $\frac{\pi }{6}$ (Ans.)